I'm currently studying vector bundles through Milnor's book Characteristic classes and he left as an exercise the definition of a quotient bundle. I tried myself to give a definition and I wanted to check if everything was right. I've read a few threads about this, namely (Smooth quotient bundle and https://mathoverflow.net/q/27401) and each of them includes an answer that is basically mine (I've found it myself seperately) but as most people on this forum giving answers are advanced mathematicians, they don't include ALL the details which is exactly what matters to me here because I'm often too lazy with details ... If you could please correct me in those details and maybe give me your opinion on the solution (mostly if it's too complicated, if it's simply wrong, if I lack understanding of a deeper notion ...) that would be wonderful ! Thanks a lot :)
My solution
Let $E$ be a $n$-real vector bundle over a base space $B$ and $E'$ a $k$-subbundle of $E$. We wish to build in a natural way a quotient bundle $E/E'$.
First of all, it is natural to me to have $E/E' = \coprod_p E_b/E'_b$ which makes sense because $E'_b$ is a linear subspace of $E_b$ for all $b$. Before looking for a topology on this disjoint union, I decided to see what I needed.
I need trivializations for $E/E'$. Let $b \in B$. Naturally, we take $b \in U \subseteq B$ a trivialazing open for both (since you can shrink) $E_b$ and $E'_b$. The trivialization gives us a base $(y_1(b),...,y_k(b)) \ (resp. \mathcal{B} =(x_1(b),...,x_n(b))$ for $E'_b \ (resp. E_b)$ with $b \in U$. Continuity/smoothness of our trivialization gives continuity/smoothness of our coordinates.
Now, the issue I encountered was that our bases weren't compatible in the sense that in the end we want a trivialization of the form $E_p/E'_p \simeq \mathbb{R}^n/\mathbb{R}^k$ but we need $E'_p = Span(x_1(b),\dots,x_k(b))$ $(*)$ (or any choice of $k$ elements of our basis). Indeed, what we want to get is the following commutative diagram :
$$\require{AMScd} \begin{CD} 0 @>>> E'_b @>>> E_b @>>> E'_b/E_b = (E'/E)_b @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> \mathbb{R}^k @>>> \mathbb{R}^n @>>> \mathbb{R}^n/\mathbb{R}^k = \mathbb{R}^{n-k} @>>> 0\end{CD}$$
Where the last arrow is obtained with the Five Lemma and is an isomorphism because all the other arrows are (it's our hypothesis).
We can see that our diagram is commutative if $(*)$ is fullfiled. Since it is not immediate, I tried to fullfil it which is done by completing our basis (up to renaming) $(y_1(b),\dots,y_k(b))$ in $\mathcal{B}' = (y_1(b),\dots,y_k(b),x_{k+1}(b),...,x_n(b))$. However, this completion might depend on $b$ but we can restrict $U$ to its intersection with the open set of $b'$ such that $\det(y_1(b'),\dots,y_k(b'),x_{k+1}(b'),\dots,x_n(b')) \neq 0$ which is non empty because it contains $b$.
Now we get an isomorphism $u(b)$ sending $\mathcal{B}$ on $\mathcal{B}'$ and another trivialization by composing our initial trivialization with $u$ ($u$ is continuous/smooth because it depends continuously/smoothly on the coordinates). With this new trivialization we achieve $(*)$ and our commutative diagram gives a trivialization of $E/E'$ in a neighbourhood of $b$. Now, let's see what naturaly topology allows our trivialization to be smooth.
Naturally one wants to chose the quotient topology on $E_b/E'_b$ and the disjoint union on $E/E'$. With those, we have the projection $E_p \to (E/E')_p$ continuous/smooth in our diagram and our isomorphisms $(E/E')_p \simeq \mathbb{R}^k$ glue in a continuous/smooth map.
That concludes my demonstration !
Please fell free to tell me anything about it to improve it (or even just correct it haha). I hope this might also help people who had the same questions and wanted to find/check some details !
Have a good day/night :)